crazyboy
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Problems with formic acid titration
I have to determine the concentration of various solutions of formic acid. On paper the easiest way to do this is by titration with sodium hydroxide
solution with phenolphthalein indicator. However, I have been having problems with this.
The procedure is as follows: ~100mg of formic acid soloution to be titrated is weighed exactly in a 100ml volumetric flask. The flask is filled to the
100ml line with distilled water and a 5ml aliquot is taken in a erlenmeyer flask, to which 3 drops of phenolphthalein is added. The solution is
titrated with 0.05M sodium hydroxide until a faint pink persists. The number of moles of sodium hydroxide are calculated and an eqimolar amount of
formic acid is assumed to be in the 5ml aliquot this is multiplied by twenty and converted to grams and compared to the actual weight of the sample
added.
The problem is that all the readings appear very high >90% when the expected values are 40-80% in one instance 101% was calculated.
The volumetric flask and pipets are washed with distilled water between each measurement, the pipets are volumetric pipets which should be highly
accurate, the buret is washed with distilled water than flushed twice with titrant. The sodium hydroxide was purchased from a reliable supplier.
We even went through lengths to prepare sodium hydroxide by titrating a HCl solution with potassium bicarbonate than titrating a freshly prepared
sodium hydroxide solution with the HCl of known concentration but to no avail.
Any ideas whats going wrong or does anyone have a better way to determine the concentration of formic acid solution? References would be great. This
has been frustrating me for the past two days.
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Magpie
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Quote: Originally posted by crazyboy  |
The procedure is as follows: ~100mg of formic acid soloution to be titrated is weighed exactly in a 100ml volumetric flask.
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100mg equals about 2 drops, right? How can you weigh this tiny amount out accurately using a 100ml volumetric flask?
Would it be possible to use a larger sample? Otherwise I don't see anything wrong with your procedure.
How much of the 0.05M NaOH does it take for the neutralization?
The single most important condition for a successful synthesis is good mixing - Nicodem
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elementcollector1
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Weighed, not measured volume. Still, that raises a good point: Why on earth would you use a 100mL volumetric flask to weigh such a tiny amount (unless
you were planning to dilute it)?
Elements Collected:52/87
Latest Acquired: Cl
Next in Line: Nd
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unionised
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If you take 100 mg and dilute it to 100 ml then take 5 ml of that solution you are titrating just 5 mg of the original acid. If it's 40%w/w then you
have 2 mg of formic acid i.e. 43µ moles of acid.
On the other hand, a litre of air will contain about 500 µg of CO2 which will dissolve in water to give about 11 µ moles of carbonic acid. Since
that's a dibasic acid it's equivalent to 22µ moles of formic acid.
A litre of air is about half as much acid as the sample you are measuring.
Dissolved CO2 in the water is a probable explanation of the error you are finding.
The simplest solution is to use more concentrated solutions- either take more of the acid, or dissolve it in less water.
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blogfast25
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I don’t see the point in titrating with 0.05 N NaOH here.
Use 0.1 N NaOH. Assume the formic acid to be 50 w%. Weigh accurately about 2.30 g of formic acid (FA) solution and dilute to 250.0 ml with DW/DIW.
Titrate 20.0 ml of this solution with standardised 0.1 N NaOH titrant solution, repeatedly.
Gram FA in your total sample, m = (Vav x c x t x 46.02538 x 12.5) g FA, with Vav the average volume of tritrant used (in L), c
nominal concentration of titrant (0.1 N), t titer of titrant solution.
% FA in solution = m / 2.30 x 100 %
| Quote: Originally posted by crazyboy |
We even went through lengths to prepare sodium hydroxide by titrating a HCl solution with potassium bicarbonate than titrating a freshly prepared
sodium hydroxide solution with the HCl of known concentration but to no avail.
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‘We even went through lengths’?? What you did was to standardise your NaOH solution, that is, no pun intended, standard procedure. To
standardise NaOH a primary standard of potassium hydrogen phthalate is commonly used. It's what I use very routinely and it works well.
| Quote: Originally posted by elementcollector1 | | Weighed, not measured volume. Still, that raises a good point: Why on earth would you use a 100mL volumetric flask to weigh such a tiny amount (unless
you were planning to dilute it)? |
It would be accurately (0.1 mg if possible) weighed in a weighing boat, preferably closed because the sample is volatile, then transferred
quantitatively into the volumetric flask, then diluted to the volumetric mark.
[Edited on 22-6-2013 by blogfast25]
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crazyboy
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The volumetric flask is tared on a laboratory balance. I use a 100ml volumetric flask because I dilute it to 100ml. It takes approximately 7.5ml of
.05M NaOH to neutralize 5ml of the solution. The amount of sample is limited so taking 2g would not be feasible. Perhaps I will try more concentrated
solutions as suggested.
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blogfast25
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0.0075 x 0.05 x 46 x (100/5) = 0.345 g formic acid in your 100 mg of sample
100 mg sample = 0.1 g
% FA = 0.345 / 0.1 x 100 % = 345 % FA!!
Your volume of titrant used seems far too high. You should be using only about 1 ml of titrant to neutralise 5 ml of the diluted solution, assuming
the 100 mg of sample contained about 50 mg FA.
Are you sure of the concentration of the titrant solution? To prepare 1 L of 0.05 N NaOH requires 1 x 0.05 x 40 = 2 g of NaOH. After that the solution
should be standardised.
Or look at it like this. You only want to sacrifice 100 mg of sample per analysis. At 50 % that's 50 mg of FA or about 1 mmol (millimol). This can be
titrated directly with 0.05 N NaOH: 0.001 mol / 0.05 mol/L x 1000 ml/L = 20 ml of 0.05 NaOH. So assuming your titrant solution is correct, you can
weigh 100 mg of sample direct into an Erlenmeyer, add about 20 ml of water and titrate with 0.05 N NaOH.
[Edited on 23-6-2013 by blogfast25]
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Magpie
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Here's another check, FYI:
MW formic acid (FA) = 46.03 g/mole
mmoles in (5/100)100mg 50% FA = (5/100)(0.5)100mg/46.03 = 0.0543 mmoles
required volume of 0.05M NaOH titrant = V
V = 0.0543 mmoles/(50 mmoles/1000mL) = 1.09mL
The single most important condition for a successful synthesis is good mixing - Nicodem
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